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13 May 2022, 05:45
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C
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Difficulty:
95%
(hard)
Question Stats:
51% (02:54) correct
49%
(02:41)
wrong
based on 120
sessions
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In a race between A and B, both start simultaneously from the same point but A runs clockwise and B runs anticlockwise. They meet for the first time at a distance of 300 meters clockwise from the starting point and for the second time at a distance of 200 meters anticlockwise from the starting point. Find the ratio of speeds of A and B, if it is known that A has not completed one full round until the second meeting.
A. 3 : 2
B. 1 : 1
C. 3 : 5
D. 1 : 3
E. Cannot be determined
Are You Up For the Challenge: 700 Level Questions
A. 3 : 2
B. 1 : 1
C. 3 : 5
D. 1 : 3
E. Cannot be determined
Are You Up For the Challenge: 700 Level Questions
19 May 2022, 22:27
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Let the circumference of the track be 'c' meters.
A travels 300 meter and B travels (c-300) meter to their first meeting.
Speed of A (SA)/Speed of B (SB) = 300/(c-300)
A then travels (c-500) meters and B travels (300+200=500) meters to their 2nd meeting
SA/SB = (c-500)/500
300/(c-300)=(c-500)/500...> d^2-00=0...> d(d-800)=0...> d=800. Therefore, after starting A and B travel for 300 meters and (800-300=500) meters respectively to their 1st meeting. Therefore, SA/SB=300/500=3/5.
ANS: C
A travels 300 meter and B travels (c-300) meter to their first meeting.
Speed of A (SA)/Speed of B (SB) = 300/(c-300)
A then travels (c-500) meters and B travels (300+200=500) meters to their 2nd meeting
SA/SB = (c-500)/500
300/(c-300)=(c-500)/500...> d^2-00=0...> d(d-800)=0...> d=800. Therefore, after starting A and B travel for 300 meters and (800-300=500) meters respectively to their 1st meeting. Therefore, SA/SB=300/500=3/5.
ANS: C
mimundertaker
Joined: 10 Jun 2025
Last visit: 23 Apr 2026
Posts: 14
Given Kudos: 64
Location: India
Schools: HEC MiM '26 ESSEC MiM '26 MiM '26
09 Oct 2025, 14:11
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a quicker way to solve I think is
if at the first meeting point:
because time for both is equal, distance is in the ration of speed which is: A/B=300/x-300
so basically,
if A travels 300 out of x, then A will travel 600 out of 2x.
let's consider the second meeting point now -
A has travelled: x-200
B has travelled: x+200
together they've travelled: x-200+x+200 = 2x
so A has travelled x-200 out of 2x.
Equate what we know from earlier: x-200=600 >>> x=800
so A/B = 300/800-300 = 3/5
if at the first meeting point:
because time for both is equal, distance is in the ration of speed which is: A/B=300/x-300
so basically,
if A travels 300 out of x, then A will travel 600 out of 2x.
let's consider the second meeting point now -
A has travelled: x-200
B has travelled: x+200
together they've travelled: x-200+x+200 = 2x
so A has travelled x-200 out of 2x.
Equate what we know from earlier: x-200=600 >>> x=800
so A/B = 300/800-300 = 3/5
Bunuel
